Question

What causes a solution to a rational equation to be an extraneous solution?

Answer 1

Extraneous solution:

An extraneous solution is a solution that arises from the solving process that is not really a solution at all

So, firstly we will solve for the equation

and then we verify each solutions by plugging them back

If denominator of rational equation becomes zero , then that solution must be extraneuous solution

For example:

$\frac{1}{x+3} +\frac{2}{x} =-\frac{3}{x(x+3)}$

we can solve for x

Multiply both sides by x(x+3)

$\frac{1}{x+3}x\left(x+3\right)+\frac{2}{x}x\left(x+3\right)=-\frac{3}{x\left(x+3\right)}x\left(x+3\right)$

now, we can simplify it

$x+2\left(x+3\right)=-3$

$x+2x+6=-3$

$3x=-9$

now, we can solve for x

$x=-3$

now, we can check whether x=-3 is extraneous solution

we will plug back x=-3 into original

$\frac{1}{-3+3} +\frac{2}{-3} =-\frac{3}{-3(-3+3)}$

$\frac{1}{0} +\frac{2}{-3} =-\frac{3}{-3(0)}$

we can see that denominator becomes 0

so, x=-3 can not be solution

so, x=-3 is extraneous solution...........Answer